TITLE: The algebraic closure of a finite field and its Galois group
QUESTION [16 upvotes]: $F$ is an extension field of a field $K$.
Let $F$  be an algebraic closure of $\mathbb{Z}_p $ ($p$ prime). Show that
$(i)$ $F$ is algebraic Galois over $\mathbb{Z}_p$
$(ii)$  The map $\alpha:F\rightarrow F$ given by $u\mapsto u^p$ is a nonidentiy $\mathbb{Z}_p$-automorphism of $F$.
$(iii)$  The subgroup $H=\langle \alpha \rangle$ is a proper subgroup of Aut$(F/\mathbb{Z}_p)$ where the fixed field is $\mathbb{Z}_p$, which is also the fixed field of Aut$(F/\mathbb{Z}_p)$ by $(i).$  
So, here is my attempt for (i). Let  $S \subset \mathbb{Z}_p[x])$ of monic polynomials of the form $x^{p^n}-x$.  Then for all $f\in S$, gcd$(f,f^{\prime})=1$(i.e, the polynomials are separble) and $F=\mathbb{Z}_p(a\in F:f(a)=0)$. So $(F/\mathbb{Z}_p)$ is Galois. I'd be glad if I could get assistance for (ii) and (iii) as well.  Thanks.
ADDED:  Attempt at (iii).   but I know that since $F/\mathbb{Z}_p$ is a finite Galois extension, its fixed field is $\mathbb{Z}_p$ hints...
The field $\mathbb{Z}_p$ must be contained in $F$. For $a\in \mathbb{Z}_p$, $\alpha(a)=a^p=a$. Thus the polynomial $x^p-x$ has $p$ zeros in $F$, namely, the elements of $\mathbb{Z}_p$. But the elements fixed under $\alpha$ are precisely the zeros in $F$ of $x^p-x$. Hence the fixed field of $\alpha$ is $\mathbb{Z}_p$ which is also the fixed field of Aut$(F/\mathbb{Z}_p)$.
Left to show that $H$ is a proper subgroup... If it helps I know that the order of $\langle \alpha \rangle $ is $n$...

REPLY [3 votes]: The last statement recommended in Arturo Magidin's answer cannot work because it is not true at all: every nontrivial element is Galois group $\mathrm{Aut}_{F_{p}}F$ has infinite order (for reference see Exercise 15 on page no. 71 of this book). Let $K$ be the union of subfield in the first chain, i.e, union of $F_{{p}^{q^{n}}}$, then it is a field. Consider a homomorphism $h$ from $K$ to $F$ given by $x\mapsto x^p$; then by extension theorem there is an $F_p$ automorphism from $F$ to $F$. Since $H$ is not a proper subgroup, it implies that $h=α^n$ for some interger $n$ ( $n \neq0$). Apply $x$ both sides where $x$ is in $F_{p^{2^{n+1}}}$, it leads to $p^{2^{n+1}}$ less than roots of equation $h(x)=α^n(x)$, a contradiction.